Problem: There are some tasks which have the following structure: you are given a model, and you can do some operations, you should use these operations to achive the goal. One way to create a new task is to use the same model and same operations, but change the goal.

Let's have a try. I have created the following task for Topcoder SRM 557 Div1-Hard: you are given n integers x1, x2, ..., xn. You are allowed to perform the assignments (as many as you want) of the following form xi ^= xj (in the original task i and j must be different, but in this task we allow i to equal j). The goal is to maximize the sum of all xi.

Now we just change the goal. You are also given n integers y1, y2, ..., yn. You should make x1, x2, ..., xn exactly equal to y1, y2, ..., yn. In other words, for each i number xi should be equal to yi.

Input

The first line contains an integer n (1 ≤ n ≤ 10000). The second line contains n integers: x1 to xn (0 ≤ xi ≤ 109). The third line contains n integers: y1 to yn (0 ≤ yi ≤ 109).

Output

If there is no solution, output -1.

If there is a solution, then in the first line output an integer m (0 ≤ m ≤ 1000000) – the number of assignments you need to perform. Then print m lines, each line should contain two integers i and j (1 ≤ i, j ≤ n), which denote assignment xi ^= xj.

If there are multiple solutions you can print any of them. We can prove that under these constraints if there exists a solution then there always exists a solution with no more than 106 operations.

Examples

Input

2
3 5
6 0


Output

2
1 2
2 2


Input

5
0 0 0 0 0
1 2 3 4 5


Output

-1


Input

3
4 5 6
1 2 3


Output

5
3 1
1 2
2 2
2 3
3 1


Input

3
1 2 3
4 5 6


Output

-1

Note

Assignment a ^= b denotes assignment a = a ^ b, where operation "^" is bitwise XOR of two integers.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int n;
int highbit(int x) {
  if (x) return highbit(x >> 1) + 1;
  return -1;
}
struct class_answer {
  int x[10005];
  int t1[1000005];
  int t2[1000005];
  int m;
  void add(int X, int Y) {
    t1[m] = X;
    t2[m] = Y;
    m++;
  }
  void print() {
    for (int i = 0; i < m; i++) printf("%d %d\n", t1[i] + 1, t2[i] + 1);
  }
  void down_print() {
    for (int i = m; i > 0; i--) printf("%d %d\n", t1[i - 1] + 1, t2[i - 1] + 1);
  }
  void onwxor(int i, int j) {
    x[i] ^= x[j];
    add(i, j);
  }
  void onwswap(int i, int j) {
    if (i != j) onwxor(i, j);
    onwxor(j, i);
    onwxor(i, j);
  }
  void doit() {
    for (int i = 0; i < min(31, n); i++) {
      for (int j = i + 1; j < n; j++) {
        if (x[i] < x[j]) {
          onwswap(i, j);
        }
        if ((x[i] ^ x[j]) < x[j]) {
          onwxor(j, i);
        }
      }
    }
    for (int i = min(30, n - 1); i >= 0; i--) {
      for (int j = 0; j < i; j++)
        if ((x[i] ^ x[j]) < x[j]) {
          onwxor(j, i);
        }
    }
  }
} answer, answer2;
bool check() {
  for (int i = 0; i < min(31, n); i++) {
    int t = 0;
    if (highbit(answer.x[i]) > highbit(answer2.x[i])) answer.onwxor(i, i);
    for (int j = i + 1; j < min(31, n); j++) {
      if ((answer.x[i] ^ answer.x[j]) <= answer2.x[i]) {
        answer.onwxor(i, j);
      }
    }
    if (answer.x[i] != answer2.x[i]) return false;
  }
  return true;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &answer.x[i]);
  for (int i = 0; i < n; i++) scanf("%d", &answer2.x[i]);
  answer.doit();
  answer2.doit();
  if (check()) {
    cout << answer.m + answer2.m << endl;
    answer.print();
    answer2.down_print();
  } else {
    cout << "-1" << endl;
  }
  return 0;
}